# How to prove that det(A)=det(B) [closed]

Let $A$ and $B$ be $n\times n$ matrices with real values and $n\geq 2$. Suppose that $A$, $B$ and $A+B$ are invertible matrices and $A^{-1}+B^{-1}=(A+B)^{-1}$. Prove that $$\det(A)=\det(B).$$

Is this true for matrix with complex values?

Thank you.

## closed as unclear what you're asking by Clement C., Omnomnomnom, DonAntonio, projectilemotion, user26857Feb 10 '17 at 22:47

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• The question in the body and in the title are different. Also, the claim in the body is false, take for example $A=I,B=2I$. – Wojowu Feb 3 '17 at 18:48
• The claim in the body it's not true even for numbers, that are special cases of 1x1 matrices. – Santropedro Feb 3 '17 at 18:50
• Maybe det(A)=det(B) is part of the hypothesis – user261263 Feb 3 '17 at 18:51
• @EugenCovaci Even then $A=B=I$ gives a counterexample. – Wojowu Feb 3 '17 at 18:51
• I think the question is supposed to be show that $A^{-1} + B^{-1}$ is invertible. I have no idea why the title doesn't match, though. – Omnomnomnom Feb 3 '17 at 18:54

The assumption is $$I=(A^{-1}+B^{-1})(A+B)=A^{-1}B+B^{-1}A+2I.$$ Subtracting $I$ from both sides gives $$0=A^{-1}B+B^{-1}A+I.$$ Left-multiplications by $A$ and, separately, by $B$ yield the two equations $$\cases{0=B+AB^{-1}A+A\\0=BA^{-1}B+A+B.}$$ Consequently $$BA^{-1}B=AB^{-1}A.$$ Because $\det$ is multiplicative, $$\det(B)^2(\det{A})^{-1} = \det(A)^2(\det{B})^{-1},$$ whence $$(\det{B})^3 = (\det{A})^3.$$

For real coefficients you may take cube roots to conclude the determinants are equal. But in any field with a nontrivial cube root of unity $\omega$ (such as the complex numbers), the result does not necessarily follow. Indeed, $$\omega + \omega^{-1}=\omega+\omega^2=-1 = (-1)^{-1} = (\omega+\omega^2)^{-1}$$ gives the counterexample $A = (\omega), B=(\omega^2)$.

• Thank you very much for your help – Majorana_2017 Feb 3 '17 at 21:02

One can reformulate your conditions in the following way $$(A^{-1} + B^{-1}) (A + B) = 1$$ $$2 + B^{-1} A + A^{-1} B = 1$$ $$B^{-1} A + A^{-1} B = -1$$

Since $B^{-1}A ~~ A^{-1}B = 1$ We can reformulate with $Q = B^{-1}A$

$$Q + Q^{-1} = -1$$ then $$Q^2 + 1 = -Q$$ Now we can bring $Q$ to Jordan Normal Form and only need to consider one Block $$J^k(\alpha)^2 = - J^k(\alpha)-1$$ $$J^k(\alpha)^2 = - J^k(\alpha + 1)$$ this will only work if blocksize $k=1$ and eigenvalue $- \alpha^2 = \alpha + 1$ So the only valid $\alpha = e^{\pm i 2/3 \pi}$

So if you want to restrict yourself to real matrices the only $A,B$ that satisfy your demand are similar to

$$A = \bigoplus_i R(2/3 \pi + \phi_i); ~~ B = \bigoplus_i R(-\phi_i )$$

Where $R$ is a Rotation Matrix

And then indeed $\det A = \det B = 1$

In the Complex case this statement does not hold. Choose e.g. $A= e^{i 2/3 \pi} I_2$ and $B = I_2$ with $I_2$ the identity matrix

• I just modified my statement, I was wrong. – Majorana_2017 Feb 3 '17 at 19:42
• Should fit your question. – the_architect Feb 3 '17 at 19:56